Solutions to the Exercises of Chapter 8

Transcription

1 8A Domains of Functions Solutions to the Eercises of Chapter 8 1 For 7 to make sense, we need 7 0or7 So the domain of f() is{ 7} For + 5 to make sense, +5 0 So the domain of g() is{ 5} For h() to make sense, both f() and g() must make sense So the domain of h() is{ 5 7} For 4 to be defined, we must have 4 0 So 4 and hence the domain of f() is{ 4} For to make sense, needs to be greater than or equal to 0 So 0 and hence the domain of g() is{ } Note that for, both f() and g() are defined For k() to make sense, we also need g() 0 So the domain of k() is { > } In the case of f(), we need both +6 0 and +6 So 6 and Hence 6 and So the domain of f() is{ 6 } Forg() to make sense, we have to have 5 0 So either +> 0 and 5 0; or +< 0 and 5 0 So + 5or< Hence the domain of g() is{ < or 5} 8B Evaluations of Limits 4 lim ( + 1)( +4) = (5)(1) = 60 5 lim 1 = +4 1 = lim 4 + = 4+= 6 7 lim +1 + = = 18 6 = 8 Notice that in the limit lim +1, the numerator goes to 4 while the denominator goes + to 0 So the ratio becomes larger and larger Because it does not close in on a finite number, there is no limit Check that lim +1 = and lim +1 = This is a limit of 0 type So we are looking for a cancellation Because 0 1 = ( + )( 4), we see that lim 1 (+)( 4) ( 4) = 7 Check that + + L Hospital s rule gives the same answer 10 This is another limit of 0 type It is solved with a cancellation: lim 0 t 1 t = 1 Check that L Hospital s rule gives the same answer lim t 1 t t t(t 1) = t 1 t 1 (t 1) 11 This is a limit of 0 type By L Hospital s rule, lim 0 1 limit can also be solved with a cancellation = Check that this

2 1 This limit is also a 0 limit that can be solved by a cancellation: 0 lim h 0 (h 5) 5 h h 0 h 10h +5 5 h h 0 h(h 10) h h 0 (h 10) = 10 Show that L Hospital s rule gives the same thing For which function f() is this limit equal to f ( 5)? 1 By rationalizing, factoring, and canceling, 81 = 81 + = ( +)=( + 9)( +) So lim 9 81 ( + 9)( + ) = 108 Check that L Hospital s rule gives the same result 9 14 By rationalizing, 1+ 1 = lim 0 = ( 1++1) (1+) = What do you get with L Hospital s rule? 15 By rationalizing, 4 s = 4 s 4+ s s 16 s = 16 s s What answer does L Hospital s rule provide? 16 The limit lim (s 16)(4+ s) = 1 4+ So lim s s 16 = 1++1 So lim 0 4 s s 16 s = 1 4+ = 1 s 8 does not eist To see this, check that lim ( ) = 1 This is so, because for <, we get < 0, so that = ( ) Show in a similar way that lim lim + does not eist = 1 Since the limit from the left is not equal to the limit from the right, 17 Because lim f() = 5 and lim g() =, we see that lim a a a 5 = 10 = 5 5 8C Continuity f() = 5 Similarly, lim g() a f() g() f() = 18 For to make sense we need only for to be non-zero By the quadratic formula, = 0 when = 1± 1+4 = 1±5 = 1 or 1 So the domain of 1 1 G() = is { 1 and 1} For + 1 to make sense, we need +1 0, or 1 For +1 to make sense we need > 1 So the domain of H() = 1 +1 is { > 1} That G() is continuous on its domain follows from the third Remark in Section 8 That H() is continuous on its domain follows from a combination of the second and third Remarks of Section 8 and the fact (see Section 85) that the composite of two continuous functions is continuous 19 That the functions c + 1 and c 1 are continuous for any constant c follows from the third Remark in Section 8 For the function f() to be continuous, its graph must be in one connected piece In view of what was already said, this will be so precisely if the graphs of c + 1 and c 1 meet when = So we need to have c +1=9c 1 So 6c = and c = 1

3 0 For (i), we need to check that the Continuity Criterion is satisfied for c = 5 Because f(5) = = = 5, we know that f(5) makes sense In view of the fact that lim f() (1 + 9) = = 5 = f(5), we now know that the Continuity c 5 Criterion is satisfied So f() is continuous at c = 5 To show that g() = +1 is continuous 1 at c = 4, we need to check the Continuity Criterion for c = 4 Because g(4) = 5 = 5,g() is defined at = 4 Since lim g() = 5 = g(4), the criterion is met So g() is continuous at c =4 1 i We know from the third Remark of Section 8 that the function 1 is continuous +1 ecept when = 1 For = 1, the function 1 is not defined so its graph has a +1 gap We need to see whether defining f( 1) = 6 closes the gap Is this the case? It will be only if lim 1 = 6 However lim 1 (+1)( 1) ( 1) = So the function f() is not continuous for = 1 Hence it is not continuous on its domain ii As in (i), the function 8 is continuous, ecept when = 4 where it is not defined 4 If lim = 6, then the definition f(4) = 6 will close the gap in the graph Because ( 4)(+) lim ( + ) = 6, the gap is indeed closed So f() is continuous at = 4 and hence for all Why are f() and g() = + eactly the same function? Let f() = + + Because f( ) = ( ) +( ) += 16+4+= 10 < 0 and f( 1)=( 1) +( 1) += +1+=1> 0, it follows from the Intermediate Value Theorem that there eists some in (, 1) such that f() = + +=0 If m and M are the minimum and maimum values of f on [ 1, 1], then m < 4 M So by the Intermediate Value Theorem, there is, for any number v with v 4, a number r between 1 and 1 such that f(r) =v Taking v = π gives us the r we need 8D Tangent Lines 4 Because g () =, the slope of the tangent to the graph at the point (0, 1) is g (0)=0 By the point-slope form of the equation of a line, the equation of the tangent line is y 1 =0( 0) or y =1 5 Because h () = 1( 1) () =, we see that h ( 1) = = By the ( 1) ( ) 9 point-slope form, the equation of the tangent line is y + 1 = ( +1) or y = Because y = 1( ) (1) =, we see that the slope of the tangent line is = = 1 ( ) ( ) (6 ) 9 By the point-slope form of the equation of a line, we get that the equation of the tangent is y = 1( 6) or y = Converting the equation y = 1 into slope-intercept form, we get y = 1 ory = 1 1 So 1 is the slope of the line Net, we need the point on the graph of f() = 1 with the property that the tangent at that point has slope 1 Because f () =, this occurs when

4 = 1 So the point is ( 1,f( 1)) = ( 1, ) The equation we are looking for is that of the line through ( 1, ) with slope 1 By the point-slope form of the equation of a line we get y ( ) = 1 ( 1 15 )ory + = 1 1 or, finally, y = For the graph of f() = to have a horizontal tangent, we need to have f () =6 6 6 = 0 By the quadraticformula, 6( 1) = 0 for = 1± 5 9 For y =6 +5 to have a tangent line of slope 4, the derivative y = must be equal to 4 But 4 = implies that = 1 and this is impossible 18 0 Start with the graph of y = and then observe that the relevant diagram is shown below: Let y = m + b be one of the two lines and let ( 1,y 1 ) and (,y ) be the two points of tangency, respectively, on the graphs of f() = + 1 and g() = 1 Observe that f ( 1 )=m = g ( ) and hence that 1 = m = So = 1 Therefore, y = 1= ( 1 ) 1= 1 1= y 1 Because y 1 = m 1 + b and y = m + b, we get y 1 = m 1 +b and y 1 = m 1 +b, and hence that b = 0 and b = 0 Because m = 1, and ( 1,y 1 ) lies on the graphs of both y = m and y = + 1, we get 1 +1=y 1 = m 1 = 1 So 1 = 1 and hence 1 = ±1 When 1 = 1, we get y 1 = 1 +1 =, = 1 and y = ( 1) 1= So the two points are (1, ) and ( 1, ) These are the points in the diagram When 1 = 1, we get y 1 =( 1) +1=, = 1, and y = 1 1= So the other two points are ( 1, ) and (1, ) 1 A reformulation of the question is this: For what point on the graph of y = 1 10 will the 4

5 tangent line hit the point (10, 5)? Let this point be ( 1,y 1 ) Because the slope of the tangent line is and the point ( 1,y 1 ) lies on it, we see that the equation of the tangent is y y 1 = 1 5 1( 1 ) Since (10, 5) must be on this line, 5 y 1 = 1 5 1(10 1 ) Since ( 1,y 1 )is also on the parabola, y 1 = Therefore, = 1 5 1(10 1 ) Multiplying by 10 gives 50 1 =0 1 1So = 0, and by the quadraticformula, 1 =10± 5 Because the car is to the left of the point (10, 5), we need to take 1 =10 5 When this is the -coordinate of the headlights, the headlights will beam in on (10, 5) 8E About Derivatives Recall that f () 0 f(+ ) f() Denoting by h this becomes f f(+h) f() () h 0 h The pattern of (i) suggests that = 5 So f f( 5+h) f( 5) ( 5) Taking f() =, h 0 h we see f( 5+h) =(h 5) and f( 5) = 5 So the limit in (i) is the derivative of f() = 1+h 1 h at = 5 Because (iii) can be rewritten as lim, we see that this limit is the derivative of f() = at = 1 Only (ii) remains A reading of Section 8 informs us h 0 that f (c) Taking c = 1, we get f (1) c derivative of f() = 9 at =1 c f() f(c) f() f(1) 1 1 It follows that lim i The domain of f() = is { 0} Now to the computation of f () : ( ) ( ) f f( + ) f() + + () 0 0 [ 1 + ] [ 0 1 ( + ) ( + ) 0 1 = 1+ [ + ( + ) ] 0 [ 1+ ( + ) The domain of f () is{ 0} It coincides with that of f() ] ] is the ii The domain of f() = 6 is { 6} The derivative is obtained by rationalizing: f f( + ) f() 6 ( + ) 6 () 0 0 ( 6 )( 6 ) ( + ) 6 ( + ) ( 6 ) ( + )+ 6 6 ( + ) (6 ) ( 6 ) ( + ) = 0 6 ( + )+ 6 6 The domain of f () is{ <6} 5

6 4 The facts to remember are these: If f () > 0 for all in an interval I, then f() is increasing over I; and if f () < 0 for all in I, then f() is decreasing over I If f () = 0, then the graph of f has a horizontal tangent Going from left to right: We see that the function whose derivative has Graph a is increasing, then suddenly decreasing, then suddenly increasing, and then suddenly decreasing again This is the pattern of Graph ii The function whose derivative has Graph b is increasing, then has a horizontal tangent, then is decreasing, has another horizontal tangent, then increases until it has another horizontal tangent, and it is decreasing thereafter This is the pattern of Graph iv Similar considerations match Graph cwith Graph iii and Graph d with Graph i 5 Because ( f g evaluated at =, it is com- g() ( ) = f ()g() f()g () f ; and at =, g() g () = f ()g() f()g () = g() puted as follows: ) () is the derivative of the quotient of f() d d ( ) f() g() ( 6)() (4)(5) 4 = 8 By the chain rule, (f(g())) = f (g()) g () Evaluating this at =, we get f (g()) g () = f () 5=( )(5) = 15 6 The graph of f() = 9 is sketched below Because makes everything positive, the graph of g() = 9 is obtained by rotating the portion of the graph of f() = 9 that is below the -ais upward as shown So the graph of g() = 9 has sharp corners at = and = Sog() is not differentiable at = and = y y The functions f() and g() coincide ecept when So g () =f () = ecept when For,g() = f() So g () = f () = The graphs of f (), and g () are sketched below: 6

7 y y The discussion in Section 85A tells us that y = c + 1 is differentiable for all no matter what c is By the same discussion, y = + d is differentiable for all 0 no matter what d is So the question is this: For which c and d do the graphs of y = c + 1 and y = + d fit together in such a way that the graph of f() that results is smooth at =4? Because the graphs need to connect when = 4 (Why?), we need to have c 4 +1= 4+d, or 16c + 1 = d + Because the connection needs to be smooth (no corner), the derivative of y = c +1 at = 4 needs to be equal to the derivative of y = + d at =4 So c 4+0 = 1 (4) 1 +0, and hence 8c = 1 So c = 1 4 d =16c 1= 16 1= 1 f() f(a) 8 This is done by rationalizing: lim a a af (a) (f() f(a))( + a) a, and by the earlier equation, ( a)( + ( + a) a) a ( ) f() f(a) = a 8F Rates of Change 9 For [4, 7] this is 6,000 4,000 = 58,000 = 19, bacteria per minute For [7, 9] it is 154,000 6,000 = ,000 = 46,000 bacteria per minute 40 For [0, 5] this is 5 70 = 5 = 7 milligrams per day, and for [0, 10] it is = 55 = milligrams per day The minus sign means that the amount is decreasing at these rates 41 If the formula T () = 500 a is to hold for all with 0 80, it must hold for = 0 So 499 = T (0) = 500 a(0) So 0a = 1 and hence a = 1 Therefore 0 T () = We now get that T (0) = 500 = 4985 and T (5) = 500 = So T (0) T (5) = 05 and hence the temperature is decreasing at a rate of 05 = 1 degrees 5 0 per inch over [0, 5] The rate of change of the temperature at any is T () = 1 degrees 0 per inch So T (0) is also equal to 1 degrees per inch 0 7

8 4 Because V (0) = 000 and V (5) = 0, the average rate at which the tank drained was 000 = 5 10 gallons/minute After 10 minutes there were V (10) = 000 ( 1 5) 10 = 1080 gallons in the tank, and after 0 minutes there were V (0) = 000 ( 1 5) 0 = 10 gallons During the time V (0) V (10) 0 10 interval [10, 0], the average rate at which the water drained was = = gallons/minute The minus means that the volume of water in the tank was decreasing Note that V (t) = 6000 ( )( ) ( 1 t = 40 1 t 5) So at t = 10 and t = 0, the rates were V (10) = 144 gallons/minute, and V (0) = 48 gallons/minute, respectively 4 The volume is equal to V = V () = When changes from to 4, the average change V (4) V () in the volume is = 4 =7 When changes from to 1, this average change 4 1 V (1) V () is = (1) =791, and when changes from to 01, the average change 1 01 V (01) V () is = (01) =709 The rate of change of V when =isv () Because V () =, this is V ()= = 7 Because the surface area of the cube is 6,V () = is one-half the surface area 44 The area of a circle of radius r is A = πr It follows that the answers to (i), (ii), and (iii) A() A() are, respectively, = 9π 4π A(5) A() =5π 1571, = π(5) π 1414, and A(1) A() = π(1) π 188 Note finally that A () =πr is the circumference of the 1 01 circle of radius r, and that A () i Because P = 800 P (50) P (00) as a function of V, this is = 4 = 0016 pounds/in V The means that the pressure is decreasing (as V increases) ii Because V = 800 dv, we see that = 800 = P dp P P Correction: In the statement of Eercise 46, change If the bodies are moving, find the rate of change of F relative to r to If the distance between the bodies is changing, find the rate of change of F relative to r Note that if one of the bodies is in a circular orbit around the other (so the distance between them is fied) then F is a constant and the derivative is zero 46 This is df dr = GmM r 8G Differentiating Functions df 47 = d (16) 16 = 1,88 48 G () =( 7) + ( +1)= = f (u) = d a u = u(1+u ) (a u )u = u(1+u +a u ) du 1+u (1+u ) (1+u ) ] 50 ds dt = d dt 51 [ t 1 (t +) = 1 t (t +)+t 1 = 1 t = u(1+a) (1+u ) (t +)+t 1 = t++t t dy d = d d (4 + +1) 1 = ( ) (4 +) = 4t+ t 8

9 5 dy d = d d ( +4 5 ) 6 ( ) 11 = [6( +4 5 ) 5 ( )]( ) 11 +( +4 5 ) 6 11( ) 10 ( ) 5 Because y = 9 4 (9 4)+ (9 4) = 9 (9 4) = (9 4) 1, we get dy d =(9 4) 1 + ( 1 ) (9 4) ( 4) = 54 F () = [5( +4+6) 4 (+4)]( +4 5 ) 1 ( +4+6) 5 [ 1 ( +4 5 ) 1 ( +0 4 )] +4 5 ( ) 1 55 Because s(t) = 4 t +1 = t +1 4, we find that t 1 t 1 ( ) [ ] ( ) s (t) = 1 t +1 4 t (t 1) (t +1)t ( ) ( = 1 t 1 4 6t = 4 t 1 (t 1) 4 t +1 (t 1) t 1 t +1 ) 4 t (t 1) 56 Because y =( + +), we get dy d =( + + )( + ) On the other hand, dy =u du and du dy = + So = dy du =u( d d du d +)=( + + )( + ), as before Finally, dy =1 = (1 + + )( + ) = ()(6)(4) = 48 d 57 i The y-coordinate of the point on the circle above is y = r Because the volume of a cylinder equals area of circular base height, we get V () =(π )(y) = π r ii We need to compute V () This is V () = π ((r ) ) (r ) 1 ( ) ( ) ( ) = π (r ) 1 (r ) =π (r ) 1 (r ) 1 ( ) r = π (r ) 1 Since neither = 0 nor = r provides a maimum (because V () = 0 in either case), the remaining possibility occurs when =r,or = r When < r, then < r,so < r, and V () > 0 Similarly, when > r, then V () < 0 It follows that V () is increasing when < r and decreasing when > r So = r gives us the maimum volume ( ) iii Because V r =π r r r = 4 πr 1 r = 4 πr 1, this is the maimum volume that an inscribed cylinder has So the ratio is = 4 πr 4 πr 1 Note: A related problem was considered by Archimedes The sphere of radius r just fits into the cylinder with base the circle of radius r and height r Archimedes had derived the epression 4 πr for the volume of a sphere of radius r, so he knew that the ratio of the volume of the cylinder to 9

10 that of the sphere was (πr )(r) 4 = πr 4 = Archimedes was evidently very proud of this achievement According to the eye-witness report of the Roman statesman Cicero, the fraction and a figure of the cylinder and the inscribed sphere were etched on Archimedes s tomb (Unfortunately, the tomb appears not to eist anymore) 58 i lim 1 1 = ii lim 1 = iii lim = (4)(9)() = By the Mean Value Theorem we know that there is a number c between 0 and 9 such that f (c) = f(9) f(0) = 1 = 4 Because f () = , we need to solve 4 =1+ 1 for Doing so, we get 1 = 1, so =, and hence = ( ) = i F () =f() + f() f () ii G () = f() 1 f ()+f()+f () iii H () =4f() +4( )(f()) f () = 4 8f () f() f() 8H Calculus of Trigonometric Functions 61 Because y = sin( 1 ), we get y = cos( 1 ) ( cos 1 )= 6 Because sin (cos 4) = [sin(cos 4)], we get y = [sin(cos(4))] cos(cos 4) ( sin 4) 4= 8(sin(cos 4))(cos(cos 4))(sin 4) 6 y = [( sin )(cos )] cos (sin )( sin ) cos = sin cos +sin cos 64 Because y = sin( 1 ), we get y = sin( 1 )+ cos( 1 ) ( ) = sin( 1 ) 1 cos( 1 ) 65 y = sec () =sec () 66 y = 5(cos +1) 6 ( sin + 1)( 1 )( +1) 1 5 sin () = +1 ( +1) 1 cos 6 ( +1) 67 y = 6(1 + sec ) 5 ( sec )(sec tan ) = 18 tan (sec )(1 + sec ) 5 68 y = sec ( ) ()+tan sec = sec ( )+tan sec 69 y = 1 (1 + tan ) 1 ( sec )= sec 1+ tan 70 lim θ π cos θ 05 θ π θ π cos θ cos π θ π = ( d d cos θ) θ= π = sin θ θ= π = sin π = 71 Because y = sec, the slope of the tangent line is sec π =4 So its equation is y = 4( π )ory =4 4π + 8I Increase and Decrease of Functions 10

11 7 Because f () =, the critical numbers are obtained by solving =0 for Doing so, we get = ±1 7 F () = ( 4) (( 4)) = 4 5 ( 4) +( 8) 1 5 = 4 5 ( 4) +( 4) 1 5 ( 4)( ) 5 It follows that the critical numbers are 0, 4, and = 8 7 = ( 4)[ 4 5 ( 4)+] 5 = 74 Note that T () =( 1) + ( 1) 1 ()=( 1) ( 1) 1 = 16( 6 16 ) So the critical numbers are 1 6, 0, and = ( 1) ( 1) 1 = 6( 1)+4 ( 1) 1 = Note: The instructions for Eercises and should be more carefully worded to ask for the values of the variable at which f has a local minimum or a local maimum rather than for the local maimum and minimum values of f Observe also that it is the understanding in this tet that a local maimum or minimum of a function cannot occur at an endpoint of the domain of the function However, the absolute maimum and minimum values can occur at such endpoints 75 Because f () = 4 + 1, the critical numbers are 4± 16 4 = 4± = 1 and 1 Take 0, 1, 6 6 and as test points Since f (0) = 1,f ( 1)= 1= 1, and f () = 5, we find that f is 4 4 increasing over the intervals (, 1) and (1, ) and decreasing over ( 1, 1) It follows that f has a local maimum value at 1 and a local minimum value at 1 76 Check that f () = =4( 4) = 4( 4)( +1) So the critical numbers are 1, 0 and 4 Take, 1, 1, and 5 to be the test points Check that f ( ) = ( 8)( 6)( 1) = 48; f ( 1)= ( 9)( 1)= 9, f (1) = 4( )() = 4, and f (5) = 0(1)6 = 10 It follows that f is increasing over ( 1, 0) and (4, ), and decreasing over (, 1) and (0, 4) So f has local minima at 1 and 4 and a local maimum at 0 77 Observe first that f() is defined only when 1 or for 1 1 Note that f () = (1 ) 1 + 1(1 ) 1 ( ) =(1 ) 1 It follows that (1 ) 1 = 1 (1 ) 1 = 1 (1 ) 1 the critical numbers are ±1 and ± 1, so they are in increasing order: 1, 1 1,, and 1 Because and 1 1, we take 08, 0, and 08 as test points Check that f ( 08) < 0, f (0) > 0, and f (08) < 0 So f is decreasing over ( 1, 1 ) and ( 1, 1) and increasing over ( 1 1, ) Notice that f has a local minimum at 1 and a local maimum at 1 78 For f() to be defined we need So 1 if >0 and 1 if <0 Observing that the second alternative is impossible, we see that the domain of f consists of the interval 0 1 Check that f () =( ) 1 + 1( ) 1 (1 ) =( ) 1 + (1 ) = ( ) 1 ( )+(1 ) = 4 + = 4( ( ) 1 ( ) 1 4 ) So the critical numbers are 0,, and 1 Because ( ) , we only need the test points 1 and 4 Check that f ( 1) > 0 and f ( 4) < Therefore f is increasing over (0, ) and decreasing over (, 1) Hence f has a local maimum 4 4 at 4 11

12 79 Check that f () =1 1 When >1, 1 < 1, so 1 > 1 and hence f () =1 1 > 0 So f is increasing for >1 Because f (1) = 0, the graph of f has a horizontal tangent at the point (1, ) It follows that f is increasing over [1, ) The verification of the inequality follows from the definition of increasing function Correction: The inequality in Eercise 80 should have two < in place of the two 80 First observe that the inequality sin β < β for 0 <α<β< π sin α α 0 <α<β< π sin So we must show that f() = sin α is equivalent to for α is a decreasing function for 0 << π > sin β β This will follow from the fact that f () < 0 for 0 << π Because f() = (sin ) 1,we get that f () = (c os) 1 + (sin )( )= cos ( ) sin cos = sin = cos sin It remains to show that sin cos >0for 0 << π Consider the function g() = sin cos with in [ ] 0, π Check that g () =cos cos ( sin ) = sin So g () > 0 for in (0, π) and hence g() is increasing over (0, π ) Because g(0) = 0, this means that g() > 0 for 0 << π The inequality β < tan β for 0 <α<β< π tan α is equivalent to < tan β for 0 <α<β< π α tan α α β So we need to show that f() = tan is an increasing function Because f() = (tan ) 1, we get f () = (sec ) 1 + (tan )( )= sec tan It remains to verify that g() = sec tan >0for 0 << π Check that g () = sec +( sec )(sec tan ) sec = sec tan >0 So g() is increasing But g(0) = 0 and therefore sec tan = g() > 0 for 0 << πsof () > 0 for 0 << π tan, and f() = is increasing as asserted 81 Check that f () =1 cos So the critical points are those with 1 cos =0, or cos = 1 A look at Figure 45 tells us that there are eactly two such in [0, π] By Section 14 and Eample 411 they are = π, 5π Take π,π, and 7π as test points By Section and Eample 411, f ( π) < 4 0,f (π) > 0 and f ( 7π)=f ( π)=f ( π ) < 0 So f() is decreasing over (0, π), increasing over ( π, 5π), and decreasing over ( 5π, π) There is a local minimum at π and a local maimum at 5π 8 Check that f () = sin + cos sin = cos Because π π, the critical numbers are π, 0, and π Take π, π, π, and π as test points Because f ( π > 0, f ( π) < 4 0, f ( π) > 0, and f ( π) < 0, the function f() is increasing on ( ) π, π 4 4, decreasing on ( π, 0), increasing on ( ) ( 0, π, and decreasing on π,π) There are local maima at π and π, and there is a local minimum at 0 8 Refer to Figure 46 of Section 44 and notice that f is not defined when = π and π Check that f () =sec tan sec = sec (1 tan ) (1 tan ) = So the critical points cos occur when 1 tan = 0 and cos = 0 Because π π, we get by consulting Chapters 14 and 44 that the critical numbers are π, π, π, and π 4π Take, π, 0, π, and 4π as

13 test points Evaluate tan at 08π, 06π, 0, 04π and 08π with a calculator and conclude that f ( 4π) > 5 0,f ( π) < 5 0,f (0) > 0,f ( π) < 0, and f ( 4π ) > 0 Thus we see that f is 5 5 increasing over ( ) ( π, π 4, decreasing over π, ) ( π 4, increasing over π, ) π 4, decreasing over ( π, ) π 4, and increasing over ( π,π) There are local maima at π and π Because f is 4 4 not defined at π and π, there are no local minima 84 The derivative is g () = sin +cos A comparison of Figures 44 and 45 shows that there is only a single with π π that satisfies sin = cos Observe that = π 4 satisfies this equality and that this is the only critical number Take π and 0 as test points to get that g() is decreasing on ( π, ) ( π 4 and increasing on π, ) π 4 So f has a local minimum at π 4 85 Check that f () =( +1) Evaluating f at the critical number 1 and at the endpoints, 5, we get f( )=,f( 1)=1, and f(5)=7 So the maimum value of f is f(5) = 7 and the minimum value is f( 1) = 1 86 The derivative is f () = 1=( 4) So the critical numbers are ± Evaluating f at the critical numbers and also at and 5, we get f( ) = 10, f( ) = 17, f() = 15, and f(5)=66 So the maimum value is f(5) = 66 and the minimum value is f() = The derivative is f () = = 6( 5 + ) By the quadraticformula, the critical numbers are 1 and Evaluating f at the required points, we get f(0) = 7, f( 1 )= 975, f() =, and f() = 16 So the maimum value of f is f() = 16 and the minimum value is f()= 88 The derivative is f () = =15 ( 1) So the critical numbers are 1, 0, and 1 Evaluating f at the required points, we get f( ) = 57, f( 1) = 1, f(0) = 1, f(1) =, and f()=55 So the maimum value is f() = 55 and the minimum value is f( ) = Check that the derivative is f () = 9 So the only critical number in [ 1, ] is 0 Evaluating the function at = 1, 0 and, we get f( 1) = 8,f(0)=, and f() = 5 So the maimum value is f(0) = and the minimum value is f() = 5 90 The upper right corner of the rectangle is the point (, a b a ) with >0 The area of the rectangle is equal to A() =() ( a b a ) =4 b a (a ) 1 The domain of A is [0,a] We are looking for the value of for which the function A() attains its maimum value Differentiating A(), we get A () =4a[ b (a ) (a ) 1 ( ) ] [ ] By taking common denominators, we get A () = 4b a = 4b(a ) The value = a a (a ) 1 a(a ) 1 can be ignored because A() = 0 in this case Notice that A () = 0 precisely when = a When < a, then < a So <a and hence A () > 0 When > a, then > a So > a, and this time A () < 0 It follows that A() is increasing to the left of = a and decreasing to the right Therefore = a gives us the maimum a we are looking for The dimensions of the maimal rectangle are: the base is = 1

14 a and the height is b a a a = b a = a b = b Its area is ab 8J More Problems from the Books of L Hospital and Agnesi 91 Let d = AB and = AE The function f() = (d ), where 0 d, has to be maimized By the product rule, f () =(d ) + (d )( 1)=(d )[d ] =(d )(d ) For =0or = d, the product (d ) = 0 is not the maimum we are looking for So only = d remains So E has to be placed at the midpoint of AB It remains to check the sign of f () and to confirm that = d actually results in a maimum For < d, we have <dand hence f () > 0 For > d, >dand this time f () < 0 Observe therefore that f() increases to the left of = d and decreases to the right So = d gives us the maimum we are looking for 9 The ratio epressed as a function of is R() = AE EB = (a+)(b +c) Note that we must CE EF (b ) have 0 <<b, so that this is the domain of R By the product and quotient rules, R () = [(b + c)+(a + )( 1)](b ) [(a + )(b + c)](b ) [(b )] Letting N() be the numerator, we get: N() = (b + c)[(b ) (a + )(b )] (a + )(b ) = (b + c)[ + b + b +a ab] (b )( + a) = (b + c)[ +a ab]+(b )( a) = (b )[ +a ab]+(b )( a)+c[ +a ab] = (b )[ +a ab a]+c( +a ab) = (b )(a ab)+c( +a ab) = (c a) +(ac + ab + ab) abc ab = (c a) +a(b + c) ab(b + c) as asserted in the hint Because 0 <<b, the denominator (b ) of R () is always positive So it remains to find the numbers with 0 <<bfor which the numerator N() =(c a) +a(b + c) ab(b + c) of R () is equal to zero i) Suppose c = a Then N() =a(b + c) ab(b + c) is zero only when = b or = b 14

15 Suppose c a By the quadraticformula, ii) Suppose c>a Because >0, = a(b + c) ± 4a (b + c) +4(c a)ab(b + c) (c a) = a(b + c) ± a(b + c)[a(b + c)+b(c a)] c a = a(b + c) ± a(b + c)c(a + b) c a 1 = a(b + c)+ ac(b + c)(a + b) c a is the only possibility A comparison of the terms a(b + c) and ac(b + c)(a + b) (square them both and use c>a) tells us that 1 > 0 as required But is 1 <b? That this is so follows by reversing the following chain of inequalities: So 1 = a(b+c)+ ac(b+c)(a+b) c a a(b+c)+ ac(b+c)(a+b) c a <b a(b + c)+ ac(b + c)(a + b) <b(c a) ac(b + c)(a + b) <bc ab + ab + ac ac(b + c)(a + b) <c (a + b) a(b + c) <c(a + b) ab + ac<ca+ cb a<c is the solution we are looking for iii) Finally, suppose c<a In this case = a(b+c) ac(b+c)(a+b) is greater than b and must c a be ruled out This is so, because a(b+c) ac(b+c)(a+b) b implies (since c a<0) that c a a(b + c) ac(b + c)(a + b) b(c a) and hence that ac(b + c)(a + b) bc ba + ab + ac = c(a + b) This is not possible because c(a + b) > 0 So again, 1 = a(b + c)+ ac(b + c)(a + b) c a is the only possibility It is not hard to show that 0 < 1 <b Are we finished? Not quite! We do not as yet know that R() = AE EB CE EF minimum at the point we found Suppose c>a Then the numerator actually has a N() =(c a) +a(b + c) ab(b + c) 15

16 of R () is a parabola that opens upward We saw in (ii) that (c a) +a(b+c) ab(b+c) has one root on the negative part of the -ais and another between 0 and b on the positive part Because the parabola has y-intercept ab(b+c) it follows that the graph of the parabola lies below the -ais from =0to 1 and above the -ais from 1 to b This implies that R () < 0 for 0 << 1 and R () > 0 for 1 <<b Therefore, R() is decreasing to the left of 1 and increasing to the right of 1 So R() has a minimum at 1 as required The cases c<aand c = a are handled in a similar way 9 To compute the length L of QH we will use the Pythagorean theorem If L is to be epressed as a function of, we need to epress QD in terms of By similar triangles, QD = CB,so DC BH QD = DC CB AD = AB = bd as required Observe that BH BH (bd ) ( ) b + d L = + d +(b + ) = d +(b + ) =(b + ) +1 ( So f() =L =(b + ) 1+ ), d where >0 Differentiating, we get ) f () = (b + ) (1+ d +(b + ) ( d ) ( ) + d (b + )d = (b + ) = ( + b)( d b) So f () = 0 only when =(d b) 1 = d b The fact that f () is not defined at = 0 can be ignored (Why?) That = d b gives us the QH of minimal length can be confirmed by noticing that f () < 0 when < d b and that f () > 0 when > d b For =(d b) 1 = d b 1,weget L = (b +(bd ) 1 ) 1+ d (bd ) This is the shortest that L can be 8K Newton s Method for Solving Equations 94 Setting f() = 1 1=0,weget 1 = 1 and hence = So = ± are the roots of 1 1 Let s see what Newton s Method gives us Note that f () = Starting with c 1 =, we get c = f() 1 f () = 1 c = f() f ( ) = 1 ) 1 c 4 = f(14167) f (14167) 16 = 1 = = =14167 = = 1414

17 Checking (with a calculator) that = we see that Newton s method has already closed in on the root to within the required four decimal place accuracy This should mean that c 5 =1414 rounded to four decimal places Let s check Because c 4 = 1414, c 5 =1414 f(1414) = f (1414) So c 5 turns out to be an approimation of that is accurate not only up to four but, in fact, up to nine decimal places 95 We get f () = + 7 Starting with c 1 = gives us c = f() f () = 8 6 =69 c 4 = 69 f(69) f (69) c 5 = 6467 f(6467) f (6467) c 6 = 6458 f(6458) f (6458) = =6467 = =6458 = = This agrees with c 5 when rounded off So the process is finished Refer to Eercises F of Chapter From the fact that f( 1) = = 0, it follows that + 1 divides Doing the division we get that + 7 7=( + 1)( 7) So c 6 must be an approimation of 7 (Why?) Because , this is indeed so 17